A zinc electrode is placed in `0.1M` solution of `ZnSO_(4)` at `25^(@)C`. Assuming salt is dissociated to the extent of `20%` at this dilution. The potential of this electrode at this temperature is `:`
`(E_(Zn^(2+)|Zn)^(@)=-0.76V)`
`a. 0.79V" ".b. -0.79V" "c. -0.81V." "d. 0.81V`

Answer

c

Explanation

`[Zn^(2+)]=0.1xx(20)/(100)=0.02M`
Half cell reaction `:`
`Zn^(2+)(0.02M)+2e^(-)rarr Zn(s)`
`E_((Zn^(2+)|Zn))=E^(c-)._((Zn^(2+)|Zn))-(0.059)/(2)log.(1)/([Zn^(2+)])`
`=-0.76V-(0.059)/(2)log.(1)/(0.02M)`
`=-0.76V-(0.059)/(2)(log50)`
`=-0.76V-(0.059)/(2)(log 5+log10)`
`=-0.76V-(0.059)/(2)(0.7+1) (log5 ~~0.7)`
`=-0.76V-(0.059)/(2)xx1.7 `(` Take`0.059~~0.06)`
`=-0.76V-0.03 xx1.7`
`=-0.76V-0.051V` ltbr. `=-0.811V`